# Dirichlet L-series and Gamma function question

Could someone help me, please, with this exercise?

Consider a sequence of complex numbers $\{a_n\}$ such that $a_n=a_m$ iff $n\cong m$ mod $q$ for some positive integer $q$.

Define the Dirichlet L-series associated to the sequence by

$$L(s)=\sum_{n=1}^{\infty} \frac{a_n}{n^s} \ \ \ \text{ for Re}(s)>1.$$

Also define $$M(x)=\sum_{m=0}^{q-1}a_{q-m} e^{mx}\ \ \ \text{ with }\ \ a_0=a_q.$$

Questions

1. How can we show that $$L(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{M(x)x^{s-1}}{e^{qx}-1}dx, \ \ \text{for Re}(s)>1 ?$$ Note: $\Gamma(s)$ is the Gamma function.
2. How does that imply $L(s)$ is continuable into the complex plane, with the only possible singularity a pole at $s=1$.

Any help is really appreciated.

• What have you tried? Do you know at least one example of this phenomenon (e.g., where every $a_n$ is 1)? – KCd Apr 1 '13 at 0:11
• Hint: Look at the analytic continuation of the Riemann zeta function. It's pretty much that same proof, except now we're allowing for the inclusion of a Dirichlet character. – Brent J Apr 1 '13 at 2:18

(1) Hint: write $$\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{M(x)x^{s-1}}{e^{qx}-1}dx = \frac{1}{\Gamma(s)} \sum_{m=0}^{q-1} \int_{0}^{\infty}\frac{a_{q-m}e^{mx}x^{s-1}}{e^{qx}} \big( 1 + e^{-qx} + e^{-2qx} + e^{-3qx} + \cdots \big) \,dx$$ and integrate term by term.
(2) Hint: in the equation you showed in part (1), write $M(x) = (M(x)-M(0))+M(0)$ and split into two integrals. The first integral should converge for any complex $s$, while the second integral will produce a pole at $s=1$.
• Your method in (2) will give you an analytic continuation to $\textrm{Re}(s)>0$ except a possible pole at $s=1$. – Sungjin Kim Apr 1 '13 at 15:54